/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 2 Find the minimal polynomial of e... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 7: Problem 2

Find the minimal polynomial of each of the following matrices. (a) \(\left(\begin{array}{ll}2 & 1 \\ 1 & 2\end{array}\right)\) (b) \(\left(\begin{array}{ll}1 & 1 \\ 0 & 1\end{array}\right)\) (c) $\left(\begin{array}{rrr}4 & -14 & 5 \\ 1 & -4 & 2 \\ 1 & -6 & 4\end{array}\right)$ (d) $\left(\begin{array}{rrr}3 & 0 & 1 \\ 2 & 2 & 2 \\ -1 & 0 & 1\end{array}\right)$

Short Answer

Expert verified
The minimal polynomials of the given matrices are: (a) \(m_A(\lambda) = (\lambda - 3)(\lambda - 1)\) (b) \(m_B(\lambda) = (1-\lambda)^2\) (c) \(m_C(\lambda) = (\lambda - 1)^3\) (d) \(m_D(\lambda) = (\lambda - 1)(\lambda - 2)^2\)

Step by step solution

01

Compute the characteristic polynomial

The characteristic polynomial is given by the determinant of \((A - \lambda I)\), where \(A\) is the matrix, \(\lambda\) is an eigenvalue, and \(I\) is the identity matrix. So for the matrix in question: \(p_A(\lambda) = \text{det} \left(\begin{array}{ll}2 - \lambda & 1 \\\ 1 & 2 - \lambda\end{array}\right) = (2 - \lambda)(2 - \lambda) - 1 = (\lambda - 3)(\lambda - 1)\)
02

Factor the characteristic polynomial

We have already factored the characteristic polynomial in Step 1: \(p_A(\lambda) = (\lambda - 3)(\lambda - 1)\)
03

Test each factor

Replace \(\lambda\) with the matrix \(A\) and check which factor(s) the matrix satisfies: \((A - 3I) = \left(\begin{array}{ll}-1 & 1 \\\ 1 & -1\end{array}\right) \neq 0\) \((A - I) = \left(\begin{array}{ll}1 & 1 \\\ 1 & 1\end{array}\right) \neq 0\) So, none of the factors satisfy the matrix.
04

Write down the minimal polynomial

Since none of the individual factors \((\lambda - 3)\) or \((\lambda - 1)\) are satisfied by the matrix \(A\), the minimal polynomial is the product of these factors: \(m_A(\lambda) = (\lambda - 3)(\lambda - 1).\) (b) \(\left(\begin{array}{ll}1 & 1 \\\ 0 & 1\end{array}\right)\)
05

Compute the characteristic polynomial

The characteristic polynomial for the matrix is: \(p_B(\lambda) = \text{det} \left(\begin{array}{ll}1-\lambda & 1 \\\ 0 & 1-\lambda\end{array}\right) = (1 - \lambda)^2\)
06

Factor the characteristic polynomial

We have already factored the characteristic polynomial in Step 1: \(p_B(\lambda) = (1 - \lambda)^2\)
07

Test each factor

Replace \(\lambda\) with the matrix \(B\) and check which factor(s) the matrix satisfies: \((B - I) = \left(\begin{array}{ll}0 & 1 \\\ 0 & 0\end{array}\right) \neq 0\) But note that \({(B - I)}^2 = \left(\begin{array}{ll}0 & 1 \\\ 0 & 0\end{array}\right)\left(\begin{array}{ll}0 & 1 \\\ 0 & 0\end{array}\right) = \left(\begin{array}{ll}0 & 0 \\\ 0 & 0\end{array}\right) = 0\)
08

Write down the minimal polynomial

Since the matrix \(B\) only satisfies the square \({(1-\lambda)}^2\), the minimal polynomial is: \(m_B(\lambda) = (1 - \lambda)^2\). I will now solve the last two parts of the exercise: (c) \(\left(\begin{array}{rrr}4 & -14 & 5 \\\ 1 & -4 & 2 \\\ 1 & -6 & 4\end{array}\right)\)
09

Compute the characteristic polynomial

The characteristic polynomial is given by the determinant of \((A - \lambda I)\), where \(A\) is the matrix, \(\lambda\) is an eigenvalue, and \(I\) is the identity matrix. So for the matrix in question: \(p_C(\lambda) = \text{det} \left(\begin{array}{rrr}4 - \lambda & -14 & 5 \\\ 1 & -4 - \lambda & 2 \\\ 1 & -6 & 4 - \lambda\end{array}\right) = (\lambda - 1)^3\)
10

Factor the characteristic polynomial

We have already factored the characteristic polynomial in Step 1: \(p_C(\lambda) = (\lambda - 1)^3\)
11

Test each factor

Replace \(\lambda\) with the matrix \(C\) and check which factor(s) the matrix satisfies: \((C - I) = \left(\begin{array}{rrr}3 & -14 & 5 \\\ 1 & -5 & 2 \\\ 1 & -6 & 3\end{array}\right) \neq 0\) However, \({(C - I)}^2 = \left(\begin{array}{lll}2 & -18 & 9 \\\ 0 & 2 & -1 \\\ 0 & 0 & 2\end{array}\right)^2 \neq 0\), and \({(C - I)}^3 = \left(\begin{array}{lll}1 & -21 & 11 \\\ 1 & 0 & -1 \\\ 1 & 0 & 1\end{array}\right)^3 = 0\)
12

Write down the minimal polynomial

Since the matrix \(C\) satisfies the cube \({(1-\lambda)}^3\) and no lower powers, the minimal polynomial is: \(m_C(\lambda) = (\lambda - 1)^3\). (d) \(\left(\begin{array}{rrr}3 & 0 & 1 \\\ 2 & 2 & 2 \\\ -1 & 0 & 1\end{array}\right)\)
13

Compute the characteristic polynomial

The characteristic polynomial for the matrix is: \(p_D(\lambda) = \text{det} \left(\begin{array}{lll}3-\lambda & 0 & 1 \\\ 2 & 2-\lambda & 2 \\\ -1 & 0 & 1-\lambda\end{array}\right) = (\lambda - 1)(\lambda - 2)^2\)
14

Factor the characteristic polynomial

We have already factored the characteristic polynomial in Step 1: \(p_D(\lambda) = (\lambda - 1)(\lambda - 2)^2\)
15

Test each factor

Replace \(\lambda\) with the matrix \(D\) and check which factor(s) the matrix satisfies: \((D - I) = \left(\begin{array}{rrr}2 & 0 & 1 \\\ 2 & 1 & 2 \\\ -1 & 0 & 0\end{array}\right) \neq 0\) \((D - 2I) = \left(\begin{array}{rrr}1 & 0 & 1 \\\ 2 & 0 & 2 \\\ -1 & 0 & -1\end{array}\right) \neq 0\) Note that \({(D - 2I)}^2 = \left(\begin{array}{lll}0 & 0 & 1 \\\ 0 & 0 & 2 \\\ 0 & 0 & 1\end{array}\right)^2 = 0\)
16

Write down the minimal polynomial

Since the matrix \(D\) only satisfies the square \({(2-\lambda)}^2\), the minimal polynomial is: \(m_D(\lambda) = (\lambda - 1)(\lambda - 2)^2\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Prove Theorem 7.5: Suppose \(S\) is an orthogonal set of nonzero vectors. Then \(S\) is linearly independent. Suppose \(S=\left\\{u_{1}, u_{2}, \ldots, u_{r}\right\\}\) and suppose \\[a_{1} u_{1}+a_{2} u_{2}+\cdots+a_{r} u_{r}=0\\] Taking the inner product of (1) with \(u_{1},\) we get \\[\begin{aligned}0 &=\left\langle 0, u_{1}\right\rangle=\left\langle a_{1} u_{1}+a_{2} u_{2}+\cdots+a_{r} u_{r}, \quad u_{1}\right\rangle \\ &=a_{1}\left\langle u_{1}, u_{1}\right\rangle+a_{2}\left\langle u_{2}, u_{1}\right\rangle+\cdots+a_{r}\left\langle u_{r}, u_{1}\right\rangle \\ &=a_{1}\left\langle u_{1}, u_{1}\right\rangle+a_{2} \cdot 0+\cdots+a_{r} \cdot 0=a_{1}\left\langle u_{1}, u_{1}\right\rangle \end{aligned}\\] Because \(u_{1} \neq 0,\) we have \(\left\langle u_{1}, u_{1}\right\rangle \neq 0 .\) Thus, \(a_{1}=0 .\) Similarly, for \(i=2, \ldots, r,\) taking the inner product of (1) with \(u_{i}\)\\[\begin{aligned}0 &=\left\langle 0, u_{i}\right\rangle=\left\langle a_{1} u_{1}+\cdots+a_{r} u_{r}, \quad u_{i}\right\rangle \\ &=a_{1}\left\langle u_{1}, u_{i}\right\rangle+\cdots+a_{i}\left\langle u_{i}, u_{i}\right\rangle+\cdots+a_{r}\left\langle u_{r}, u_{i}\right\rangle=a_{i}\left\langle u_{i}, u_{i}\right\rangle\end{aligned}\\] But \(\left\langle u_{i}, u_{i}\right\rangle \neq 0,\) and hence, every \(a_{i}=0 .\) Thus, \(S\) is linearly independent.

Prove each of the following: (a) \(P\) is orthogonal if and only if \(P^{T}\) is orthogonal. (b) If \(P\) is orthogonal, then \(P^{-1}\) is orthogonal (c) If \(P\) and \(Q\) are orthogonal, then \(P Q\) is orthogonal. (a) We have \(\left(P^{T}\right)^{T}=P\). Thus, \(P\) is orthogonal if and only if \(P P^{T}=I\) if and only if \(P^{T T} P^{T}=I\) if and only if \(P^{T}\) is orthogonal (b) We have \(P^{T}=P^{-1}\), because \(P\) is orthogonal. Thus, by part (a), \(P^{-1}\) is orthogonal. (c) We have \(P^{T}=P^{-1}\) and \(Q^{T}=Q^{-1} .\) Thus, \((P Q)(P Q)^{T}=P Q Q^{T} P^{T}=P Q Q^{-1} P^{-1}=I .\) Therefore \((P Q)^{T}=(P Q)^{-1},\) and so \(P Q\) is orthogonal

Show that each of the following is not an inner product on \(\mathbf{R}^{3}\), where \(u=\left(x_{1}, x_{2}, x_{3}\right)\) and \(v=\left(y_{1}, y_{2}, y_{3}\right)\) (a) \(\quad\langle u, v\rangle=x_{1} y_{1}+x_{2} y_{2}\) (b) \(\quad\langle u, v\rangle=x_{1} y_{2} x_{3}+y_{1} x_{2} y_{3}\)

Prove (a) \(\|\cdot\|_{1}\) is a norm on \(\mathbf{R}^{n}\). (b) \(\|\cdot\|_{\infty}\) is a norm on \(\mathbf{R}^{n}\).

Let \(\mathrm{S}\) consist of the following vectors in \(\mathbf{R}^{4}\) : \\[u_{1}=(1,1,0,-1), u_{2}=(1,2,1,3), u_{3}=(1,1,-9,2), u_{4}=(16,-13,1,3)\\] (a) Show that \(S\) is orthogonal and a basis of \(\mathbf{R}^{4}\). (b) Find the coordinates of an arbitrary vector \(v=(a, b, c, d)\) in \(\mathbf{R}^{4}\) relative to the basis \(S\) (a) Compute \\[\begin{array}{lll} u_{1} \cdot u_{2}=1+2+0-3=0, & u_{1} \cdot u_{3}=1+1+0-2=0, & u_{1} \cdot u_{4}=16-13+0-3=0 \\ u_{2} \cdot u_{3}=1+2-9+6=0, & u_{2} \cdot u_{4}=16-26+1+9=0, & u_{3} \cdot u_{4}=16-13-9+6=0 \end{array}\\] Thus, \(S\) is orthogonal, and \(S\) is linearly independent. Accordingly, \(S\) is a basis for \(\mathbf{R}^{4}\) because any four linearly independent vectors form a basis of \(\mathbf{R}^{4}\) (b) Because \(S\) is orthogonal, we need only find the Fourier coefficients of \(v\) with respect to the basis vectors, as in Theorem \(7.7 .\) Thus, \\[\begin{array}{ll}k_{1}=\frac{\left\langle v, u_{1}\right\rangle}{\left\langle u_{1}, u_{1}\right\rangle}=\frac{a+b-d}{3}, & k_{3}=\frac{\left\langle v, u_{3}\right\rangle}{\left\langle u_{3}, u_{3}\right\rangle}=\frac{a+b-9 c+2 d}{87} \\ k_{2}=\frac{\left\langle v, u_{2}\right\rangle}{\left\langle u_{2}, u_{2}\right\rangle}=\frac{a+2 b+c+3 d}{15}, & k_{4}=\frac{\left\langle v, u_{4}\right\rangle}{\left\langle u_{4}, u_{4}\right\rangle}=\frac{16 a-13 b+c+3 d}{435}\end{array}\\] are the coordinates of \(v\) with respect to the basis \(S\).
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Mechanics Maths

Read Explanation

Geometry

Read Explanation

Statistics

Read Explanation

Pure Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.